chapter 19 sound waves molecules obey shm s(x,t)=s m cos(kx- t) leads to variations in density and...
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![Page 1: Chapter 19 Sound Waves Molecules obey SHM s(x,t)=s m cos(kx- t) leads to variations in density and pressure in the air pressure is large where density](https://reader038.vdocuments.site/reader038/viewer/2022103005/56649d225503460f949f8b25/html5/thumbnails/1.jpg)
Chapter 19 Sound Waves• Molecules obey SHM
s(x,t)=smcos(kx-t)
• leads to variations in density and pressure in the air
• pressure is large where density is large but displacement s(x,t) is small
• hence pressure(or density) wave• is 900 out of phase with displacement
• p=p0 + pm cos(kx-t-/2) = p0 + pm sin(kx-t) (p is change in pressure relative to equilibrium)
= 0 + m sin(kx-t) “rho”
m = pm (0/B)
• B is bulk modulus
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Speed of Sound• B=-p/(V/V)
• Water: B~2.2x109N/m2 Air: B~ 105 N/m2
• gases are more compressible (B is smaller)
• speed of sound in a fluid depends on the density and the elasticity-> i.e. the medium
• v = (B/0)1/2 wave speed in a fluid
• vair ~ 343 m/s at room temperature
• vwater ~ 1482 m/s ( B is larger!)
• vsteel ~ 5941 m/s
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Intensity Level and Loudness• I=Pav/area
• area of sphere = 4r2 => I 1/r2
• sensation of loudness is not directly proportional to intensity
• varies as log(I) eg. log(100)=2log(10)
• define sound level SL=10 log(I/I0) in decibels (dB)
• I0 = 10-12 W/m2 is the hearing threshold
• hence for I=I0 , SL= 10 log(1) = 0
• pain threshold is I= 1 W/m2 or SL=10log(1012)=120 dB
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Problem • When a pin of mass 0.1 g is dropped from a height of 1 m, 0.05% of
its energy is converted into a sound pulse with a duration of 0.1 s.
• (a) Estimate the range at which the dropped pin can be heard if the minimum audible intensity is 10 -11 W/m2 .
• (b) Your result in (a) is much too large in practice because of background noise. If you assume that the intensity must be at least 10-8 W/m2 for the sound to be heard, estimate the range at which the dropped pin can be heard. (In both parts, assume that the intensity is P/4r2 .)
• (a) Sound energy is 5 x 10-4 (mgh) = 5 x 10-4 (1m)(10-4 kg)(9.8 m/s2 ) = 4.9 x 10-7 J
• Pav = E/t = 4.9 x10-6 W = 4r2 x10-11 W => r ~ 200 m.
•
• (b) r ~ 200/ (1000)1/2 = 6.24 m.
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Problem • The equation I = Pav / 4r2 is predicated on the assumption that the transmitting
medium does not absorb any energy.
• It is known that absorption of sound by dry air results in a decrease of intensity of approximately 8 dB/km.
• The intensity of sound at a distance of 120 m from a jet engine is 130 dB.
• Find the intensity at 2.4 km from the jet engine (a) assuming no absorption of sound by air, and (b) assuming a diminution of 8 dB/km.(Assume that the sound radiates uniformly in all directions.)
• (a) SL = 10 log(I1/I0)-10 log (I2/I0)= 10 log(I1/I2) = 10log(r22/r1
2)=20log(r2/r1)
• 20 log (20) = 26; SL at 2.4 km = (130 - 26) dB = 104 dB
• (b) Subtract 2.28 x 8 dB from result of (a)
• SL= (104 - 18.2) dB = 85.8 dB
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ripple
Interference of Sound Waves
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Interference andPhase Difference
• Consider shifting one of the waves by a distance relative to the other
• they add up constructively • corresponds to a phase shift by = 2• constructive interference for = m2 • or path difference L= m • a shift by = corresponds to a shift of one of
the waves by /2• destructive interference for = (2m+1) • or path difference L= (2m+1) /2
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Musical Sounds
• oscillating strings• membranes• air columns• standing wave patterns correspond to
resonances• large amplitude oscillations push surrounding
air and generate sound waves at the same frequency
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Pressure waves in open pipe
Pressure waves in pipeclosed at one end